Question

Let $\overrightarrow{q}$ and $\overrightarrow{r}$ be non-collinear vectors. If $\overrightarrow{p}$ is a vector such that $\overrightarrow{p}.\left(\overrightarrow{q}+\overrightarrow{r}\right)=4$ and $\overrightarrow{p}\times \left(\overrightarrow{q}\times \overrightarrow{r}\right)=\left(x^2-2x+6\right)\overrightarrow{q}+\left(siny\right)\overrightarrow{r}$ , then $\left(x,y\right)$ lies on the line

• A. $x+y=0$
• B. $x-y=0$
• C. $x=1$
• D. $y=\pi$

Answer 1

Answer: (C)

$\left(\overrightarrow{p}.\overrightarrow{r}\right)\overrightarrow{q}-\left(\overrightarrow{p}.\overrightarrow{q}\right)\overrightarrow{r}=\left(x^2-2x+6\right)\overrightarrow{q}+siny\left(\overrightarrow{r}\right)$
on comparison, we get,
$\overrightarrow{p}.\overrightarrow{r}=\left(x^2-2x+6\right)$ and
$-\overrightarrow{p}.\overrightarrow{q}=siny$
$\because \overrightarrow{p}.\overrightarrow{q}+\overrightarrow{p}.\overrightarrow{r}=4$ (given)
$\Rightarrow \left(x^2-2x+6\right)-siny=4$
$\Rightarrow {\left(x-1\right)}^2+1=siny$
$\because {\left(x-1\right)}^2+1\ge 1,siny\le 1$
$\therefore$ Both sides are equal for $x=1,siny=1$